Question

an airline is trying two new boarding procedures, option 1 and option 2, to load passengers onto their long beach (lgb) to san francisco (sfo) flights. since option 1 has more automation, the airline suspects that the mean option 1 loading time is less than the mean option 2 loading time. to see if this is true, the airline selects a random sample of 210 flights from lgb to sfo using option 1 and records their loading times. the sample mean is found to be 17.7 minutes, with a sample standard deviation of 5.3 minutes. they also select an independent random sample of 295 flights from lgb to sfo using option 2 and record their loading times. the sample mean is found to be 18.3 minutes, with a sample standard deviation of 4.5 minutes. since the sample sizes are quite large, it is assumed that the population standard deviation of the loading times using option 1 and the loading times using option 2 can be estimated to be the sample standard - deviation values given above. at the 0.05 level of significance, is there sufficient evidence to support the claim that the mean option 1 loading time, $mu_1$, is less than the mean option 2 loading time, $mu_2$, for the airlines flights from lgb to sfo? perform a one - tailed test. then complete the parts below. carry your intermediate computations to at least three decimal places. (if necessary, consult a list of formulas.) (a) state the null hypothesis $h_0$ and the alternative hypothesis $h_1$. $h_0$: $square$ $h_1$: $square$ (b) determine the type of test statistic to use. select (c) find the value of the test statistic. (round to three or more decimal places.) $square$ (d) find the p - value. (round to three or more decimal places.) $square$ (e) can we support the claim that the mean option 1 loading time is less than the mean option 2 loading time for the airlines flights from lgb to sfo? oyes ono

Explanation:

Step1: State hypotheses

The null hypothesis $H_0$ is that the mean of Option 1 is greater than or equal to the mean of Option 2, i.e., $H_0:\mu_1\geq\mu_2$. The alternative hypothesis $H_1$ is that the mean of Option 1 is less than the mean of Option 2, i.e., $H_1:\mu_1 < \mu_2$.

Step2: Determine test - statistic type

Since the sample sizes $n_1 = 210$ and $n_2=295$ are large ($n_1\geq30$ and $n_2\geq30$) and the population standard - deviations are estimated by sample standard - deviations, we use the z - test statistic for two independent samples. The formula for the z - test statistic is $z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$, under $H_0$, $\mu_1-\mu_2 = 0$.

Step3: Calculate the test - statistic value

We are given $\bar{x}_1 = 17.7$, $s_1 = 5.3$, $n_1 = 210$, $\bar{x}_2 = 18.3$, $s_2 = 4.5$, $n_2 = 295$.
\[

$$\begin{align*} z&=\frac{(17.7 - 18.3)-0}{\sqrt{\frac{5.3^2}{210}+\frac{4.5^2}{295}}}\\ &=\frac{- 0.6}{\sqrt{\frac{28.09}{210}+\frac{20.25}{295}}}\\ &=\frac{-0.6}{\sqrt{0.133762+0.068644}}\\ &=\frac{-0.6}{\sqrt{0.202406}}\\ &=\frac{-0.6}{0.4499}\\ &\approx - 1.333 \end{align*}$$

\]

Step4: Calculate the p - value

Since this is a left - tailed test, the p - value is $P(Z < - 1.333)$. Using the standard normal distribution table or a calculator, $P(Z < - 1.333)=0.091$.

Step5: Make a decision

The level of significance $\alpha = 0.05$. Since the p - value ($0.091$) is greater than $\alpha(0.05)$, we fail to reject the null hypothesis.

Answer:

(a) $H_0:\mu_1\geq\mu_2$, $H_1:\mu_1 < \mu_2$
(b) z - test
(c) $-1.333$
(d) $0.091$
(e) No